Partial fraction expansion of inverse differential operator I don't find it obvious that an inverse differential operator can be decomposed into partial fractions like ordinary polynomials and the result obtained is same as if the whole operator is operated on a function.
The book says:-
1)If $$y_{p} = \frac{1}{P(D)}Q(D)$$, then the inverse operator can also be expanded into partial fractions just as if it were an ordinary polynomial.
2)Applying each member of a partial fraction expansion of $$\frac{1}{P(D)}$$ to $$Q(x)$$ and adding the results will give the same answer as will applying the orignal operator to Q.
I don't see why this is true. 
Additional information :- 
P(D) is a polynomial in D, the differential operator, applied on y.
The original equation is : $$P(D)y = Q(x)$$ and we seek to find a particular solution of this equation.
Example : 
$$
y_{p} =\frac{1}{(D-1)(D-2)}Q(x)  = \frac{-1}{(D-1)}Q(x) +\frac{1}{(D-2)}Q(x)$$
From Elementary Differential Equations - Morris Tenenbaum
and From the same book
 A: Obvious or not, it's true. Except of course writing the inverse of an operator as a reciprocal is hideous.
Consider an example: $$\frac{2}{t^2-1}=\frac{1}{t-1}-\frac{1}{t+1}.$$
We want to verify on this basis that $$2(D^2-I)^{-1}=(D-I)^{-1}-(D+I)^{-1}.$$ There's no problem with the factorization $D^2-I=(D-I)(D+I)=(D+I)(D-I)$. So
$$((D-I)^{-1}-(D+I)^{-1})\frac12(D^2-I)=\frac12((D+I)-(D-I))=I,$$which says that $(D-I)^{-1}-(D+I)^{-1}$ is the inverse of $\frac12(D^2-I)$.
Of course that's a little "informal", to put it charitably, since these things are not even invertible, having non-trivial  nullspaces. If you were to apply this to solving $\frac12(y''-y)=f$, you'd want to say the general solution is the general  solution to  $y'-y=f$, including the "$c_1$", minus the general solution to $y'+y=f$, including the "$c_2$".
One could clean up the math a bit by calling $P(D)^{-1}$ a one-sided inverse, and/or regarding $P(D)^{-1}f$ as a set of functions. I'm not going to worry about that - it gives the right answer when I do an example, and what's above is good enough to convince me that's  not just a coincidence.
Edit: I worried about it. This may be a better way to put the argument: Say $g_1=(D+1)^{-1}f$ and $g_2=(D-I)^{-1}f$, so $(D+I)g_1=f=(D-I)g_2$. Then $$\frac12(D^2-I)(g_1-g_2)=\frac12(D-I)(D+I)(g_2-g_1)=\frac12((D+I)f-(D-I)f)=f.$$
Note Although what's above is correct, I don't see the point to it in solving differential equations - seems to me just using $(D^2-I)^{-1}=(D-I)^{-1}(D+I)^{-1}$ is simpler, no need for the partial fractions, and also easier to justify. There's nothing unclean about the following:
Note that $y''-y=z'-z$, if $z=y'+y$. So to solve $y''-y=f$, first solve $z'-z=f$ and then solve $y'+y=z$.
